In the figure, a circle touches all the four sides of a quadrilateral ABCD whose sides AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm. The length of AD is:

Answer: Option B

We know that AB + CD = BC + AD
6.5 + 5.3 = 5.4 + AD
11.8 = 5.4 + AD
AD = 11.8 - 5.4
AD = 6.4 cm
Note:-

AB + CD = (AB + PB) + (DR + RC) . . . . . . (i)
The length of tangent drawn from an external point to a circle are equal.
AB = CD, AS = BQ, CQ = RC, DR = DS
From equation (i)
AB + CD = AS + BQ + DS + CQ
= (AS + DS) + (BQ + BC)
$$\boxed{\text{AB}+\text{CD}=\text{AD}+\text{BC}}$$
If a circle is inscribed inside a quadrilateral then the sum of a pair opposite side equal to the sum of another pair of opposite sides.
i.e. AB + CD = AD + BC